What is the sum from n=0 to infinity of 1/((3n+1)^3) ?

The sum from n=0 to infinity of 1/((3n+1)^3) is converges

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sum from n=0 to infinity of 1/((3n+1)^3)

n = 0 \sum \infty \frac{1}{( 3 n + 1 ) ^{3}}

converges

Solution steps

n = 0 \sum \infty \frac{1}{( 3 n + 1 ) ^{3}}

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty (- \frac{6}{7})^{n}

n = 1 \sum \infty 90 (0.1)^{n}

i = 1 \sum \infty \frac{1}{i} - \frac{1}{i + 1}

n = 1 \sum \infty \frac{1}{5 + e ^{- n}}

m = 1 \sum \infty \frac{10}{m ^{8} + 3}

What is the sum from n=0 to infinity of 1/((3n+1)^3) ?
The sum from n=0 to infinity of 1/((3n+1)^3) is converges